10 1. Simple Examples of Propagation
The proof also yields finite speed of propagation of signals. Define
λmin(t, x) and λmax(t, x) to the smallest and largest eigenvalues of A(t, x).
Then the functions λ are uniformly Lipschitzean on [0,T]×R. To prove this,
observe that equation (1.1.9) shows that the diagonal elements cj(t, x) of the
right-hand side are the eigenvalues of A. Their expression by the left-hand
side shows that their partial derivatives of first order (in fact any order)
are bounded on [0,T] × R. Therefore the cj are uniformly Lipschitzean on
[0,T] × R. It follows that λmin := minj {cj} is uniformly Lipschitzean.
The characteristics have speeds bounded below by λmin and above by
λmax. The next result shows that these are lower and upper bounds, respec-
tively, for the speeds of propagation of signals.
Corollary 1.1.3. Suppose that −∞ xl xr and γl (resp. γr) is the
integral curve of ∂t + λmin(t, x)∂x (resp. ∂t + λmax(t, x)∂x) passing through
xl (resp. xr). Denote by Q the four sided region in 0 t T bounded on
the left by γl and the right by γr. If g is supported in [xl,xr] and f
is supported in Q, then for 0 t T the solution u is supported in Q.
Proof. The explicit formulas of the method of characteristics show that the
approximate solutions
are supported in Q. Passing to the limit proves
the result.
Consider next the case of f = 0 and g
whose restrictions to
]−∞,x[ and ]x, ∞[ are each smooth with uniformly bounded derivatives of
every order. Such a function is called piecewise smooth.
The simplest case is that of an operator ∂t + c(t, x)∂x. Denote by γ the
characteristic through x. The values of u to the left of γ are determined by
g to the left of x. Choose a ˜ g
which agrees with g to the left and
has bounded derivatives of all orders. The solution ˜ u then agrees with u to
the left of γ and ˜ u has bounded partials of all orders for 0 t T. An
analogous argument works for the right-hand side, and one sees that u is
in the decomposition of [0,T] × R into two pieces by γ.
Suppose now that A satisfies Hypothesis 1.1.2, and for all (t, x) [0,T]×
R has N distinct real eigenvalues ordered so that cj cj+1. Denote by γj
the corresponding characteristics through x. Define open wedges,
W1 := (t, x) : 0 t T, −∞ x γ1(t) ,
WN+1 := (t, x) : 0 t T, γN (t) x ,
and for 2 j N,
Wj := (t, x) : 0 t T, γj−1(t) x γj(t) .
They decompose [0,T] × R into pie shaped wedges with vertex at (0,x) and
numbering from left to right.
Previous Page Next Page